What is the pH for the potassium propionate solution at the equivalence point?

Answer 1

The pH is determined by the hydrolysis of the propionate ion, which I will call Pr^-
Pr^- + HOH ==> HPr + OH^-
Set up an ICE chart and substitute into the following Kb expression.

Kb = (Kw/Ka) = (HPr)(OH^-)/(Pr^-)
Kw you know
Ka you know
(HPr) = (OH^-) = x
(Pr^-) = you don't have that in the problem but it is the concn of the salt at that point.
Solve for x, I would then convert to pOH, then subtract from 14 to obtain pH.